Fibonacci Numbers F n = F n-1 + F n-2 F 0 =0, F 1 =1 – 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 … Straightforward recursive procedure is slow! Why? How slow? Lets draw the recursion tree

Fibonacci Numbers (2) We keep calculating the same value over and over! F(6) = 8 F(5) F(4) F(3) F(1) F(2) F(0) F(1) F(2) F(0) F(3) F(1) F(2) F(0) F(1) F(4) F(3) F(1) F(2) F(0) F(1) F(2) F(0)

Fibonacci Numbers (3) How many summations are there? Golden ratio Thus F n 1.6 n Our recursion tree has only 0s and 1s as leaves, thus we have 1.6 n summations Running time is exponential!

Fibonacci Numbers (4) We can calculate F n in linear time by remembering solutions to the solved subproblems – dynamic programming Compute solution in a bottom-up fashion Trade space for time! – In this case, only two values need to be remembered at any time (probably less than the depth of your recursion stack!) Fibonacci(n) F 0 0 F 1 1 for i 1 to n do F i F i-1 + F i-2 Fibonacci(n) F 0 0 F 1 1 for i 1 to n do F i F i-1 + F i-2

Optimization Problems We have to choose one solution out of many – a one with the optimal (minimum or maximum) value. A solution exhibits a structure – It consists of a string of choices that were made – what choices have to be made to arrive at an optimal solution? The algorithms computes the optimal value plus, if needed, the optimal solution

Two matrices, A – n m matrix and B – m k matrix, can be multiplied to get C with dimensions n k, using nmk scalar multiplications Problem: Compute a product of many matrices efficiently Matrix multiplication is associative – (AB)C = A(BC) Multiplying Matrices

Multiplying Matrices (2) The parenthesization matters Consider A B C D, where – A is 30 1, B is 1 40, C is 40 10, D is Costs: – (AB)C)D = – (AB)(CD) = – A((BC)D) = 1400 We need to optimally parenthesize

Multiplying Matrices (3) Let M(i,j) be the minimum number of multiplications necessary to compute Key observations – The outermost parenthesis partitions the chain of matrices (i,j) at some k, (i k<j): (A i … A k )(A k+1 … A j ) – The optimal parenthesization of matrices (i,j) has optimal parenthesizations on either side of k: for matrices (i,k) and (k+1,j)

Multiplying Matrices (4) We try out all possible k. Recurrence: A direct recursive implementation is exponential – there is a lot of duplicated work (why?) But there are only different subproblems (i,j), where 1 i j n

Multiplying Matrices (5) Thus, it requires only (n 2 ) space to store the optimal cost M(i,j) for each of the subproblems: half of a 2d array M[1..n,1..n] Matrix-Chain-Order(d 0 …d n ) 1 for i 1 to n do 2 M[i,i] 3 for l 2 to n do 4 for i 1 to n-l+1 do 5 j i+l-1 6 M[i,j] for k i to j-l do 8 q M[i,k]+M[k+1,j]+d i-1 d k d j 9 if q < M[i,j] then 10 M[i,j] q 11 c[i,j] k 12return M, c Matrix-Chain-Order(d 0 …d n ) 1 for i 1 to n do 2 M[i,i] 3 for l 2 to n do 4 for i 1 to n-l+1 do 5 j i+l-1 6 M[i,j] for k i to j-l do 8 q M[i,k]+M[k+1,j]+d i-1 d k d j 9 if q < M[i,j] then 10 M[i,j] q 11 c[i,j] k 12return M, c

Multiplying Matrices (6) After execution: M[1,n] contains the value of the optimal solution and c contains optimal subdivisions (choices of k) of any subproblem into two subsubproblems A simple recursive algorithm Print-Optimal- Parents(c, i, j) can be used to reconstruct an optimal parenthesization Let us run the algorithm on d = [10, 20, 3, 5, 30]

Multiplying Matrices (7) Running time – It is easy to see that it is O(n 3 ) – It turns out, it is also (n 3 ) From exponential time to polynomial

Memoization If we still like recursion very much, we can structure our algorithm as a recursive algorithm: – Initialize all M elements to and call Lookup-Chain(d, i, j) Lookup-Chain(d,i,j) 1 if M[i,j] < then 2 return M[i,j] 3 if i=j then 4 M[i,j] 0 5else for k i to j-1 do 6 q Lookup-Chain(d,i,k)+ Lookup-Chain(d,k+1,j)+d i-1 d k d j 7 if q < M[i,j] then 8 M[i,j] q 9return M[i,j] Lookup-Chain(d,i,j) 1 if M[i,j] < then 2 return M[i,j] 3 if i=j then 4 M[i,j] 0 5else for k i to j-1 do 6 q Lookup-Chain(d,i,k)+ Lookup-Chain(d,k+1,j)+d i-1 d k d j 7 if q < M[i,j] then 8 M[i,j] q 9return M[i,j]

Dynamic Programming In general, to apply dynamic programming, we have to address a number of issues: – 1. Show optimal substructure – an optimal solution to the problem contains within it optimal solutions to sub- problems Solution to a problem: – Making a choice out of a number of possibilities (look what possible choices there can be) – Solving one or more sub-problems that are the result of a choice (characterize the space of sub-problems) Show that solutions to sub-problems must themselves be optimal for the whole solution to be optimal (use cut-and-paste argument)

Dynamic Programming (2) – 2. Write a recurrence for the value of an optimal solution M opt = Min over all choices k {(Sum of M opt of all sub-problems, resulting from choice k) + (the cost associated with making the choice k)} – Show that the number of different instances of sub- problems is bounded by a polynomial

Dynamic Programming (3) – 3. Compute the value of an optimal solution in a bottom-up fashion, so that you always have the necessary sub-results pre-computed (or use memoization) – See if it is possible to reduce the space requirements, by forgetting solutions to sub- problems that will not be used any more – 4. Construct an optimal solution from computed information (which records a sequence of choices made that lead to an optimal solution)

Longest Common Subsequence Two text strings are given: X and Y There is a need to quantify how similar they are: – Comparing DNA sequences in studies of evolution of different species – Spell checkers One of the measures of similarity is the length of a Longest Common Subsequence (LCS)

LCS: Definition Z is a subsequence of X, if it is possible to generate Z by skipping some (possibly none) characters from X For example: X =ACGGTTA, Y=CGTAT, LCS(X,Y) = CGTA or CGTT To solve LCS problem we have to find skips that generate LCS(X,Y) from X, and skips that generate LCS(X,Y) from Y

LCS: Optimal Substructure We make Z to be empty and proceed from the ends of X m =x 1 x 2 …x m and Y n =y 1 y 2 …y n – If x m =y n, append this symbol to the beginning of Z, and find optimally LCS(X m-1, Y n-1 ) – If x m y n, Skip either a letter from X or a letter from Y Decide which decision to do by comparing LCS(X m, Y n-1 ) and LCS(X m-1, Y n ) – Cut-and-paste argument

LCS: Recurence The algorithm could be easily extended by allowing more editing operations in addition to copying and skipping (e.g., changing a letter) Let c[i,j] = LCS(X i, Y j ) Observe: conditions in the problem restrict sub- problems (What is the total number of sub- problems?)

LCS: Compute the Optimum LCS-Length(X, Y, m, n) 1 for i 1 to m do 2 c[i,0] 3 for j 0 to n do 4 c[0,j] 5 for i 1 to m do 6 for j 1 to n do 7 if x i = y j then 8 c[i,j] c[i-1,j-1]+1 9 b[i,j] copy 10 else if c[i-1,j] c[i,j-1] then 11 c[i,j] c[i-1,j] 12 b[i,j] skipx 13 else 14 c[i,j] c[i,j-1] 15 b[i,j] skipy 16return c, b LCS-Length(X, Y, m, n) 1 for i 1 to m do 2 c[i,0] 3 for j 0 to n do 4 c[0,j] 5 for i 1 to m do 6 for j 1 to n do 7 if x i = y j then 8 c[i,j] c[i-1,j-1]+1 9 b[i,j] copy 10 else if c[i-1,j] c[i,j-1] then 11 c[i,j] c[i-1,j] 12 b[i,j] skipx 13 else 14 c[i,j] c[i,j-1] 15 b[i,j] skipy 16return c, b

LCS: Example Lets run: X =ACGGTTA, Y=CGTAT How much can we reduce our space requirements, if we do not need to reconstruct LCS?

February 8, Shortest Path Generalize distance to weighted setting Digraph G = (V,E) with weight function W: E R (assigning real values to edges) Weight of path p = v 1 v 2 … v k is Shortest path = a path of the minimum weight Applications – static/dynamic network routing – robot motion planning – map/route generation in traffic

February 8, Shortest-Path Problems Shortest-Path problems – Single-source (single-destination). Find a shortest path from a given source (vertex s) to each of the vertices. The topic of this lecture. – Single-pair. Given two vertices, find a shortest path between them. Solution to single-source problem solves this problem efficiently, too. – All-pairs. Find shortest-paths for every pair of vertices. Dynamic programming algorithm. – Unweighted shortest-paths – BFS.

February 8, Optimal Substructure Theorem: subpaths of shortest paths are shortest paths Proof (cut and paste) – if some subpath were not the shortest path, one could substitute the shorter subpath and create a shorter total path

February 8, Triangle Inequality Definition – (u,v) weight of a shortest path from u to v Theorem – (u,v) (u,x) + (x,v) for any x Proof – shortest path u Î v is no longer than any other path u Î v – in particular, the path concatenating the shortest path u Î x with the shortest path x Î v

February 8, Negative Weights and Cycles? Negative edges are OK, as long as there are no negative weight cycles (otherwise paths with arbitrary small lengths would be possible) Shortest-paths can have no cycles (otherwise we could improve them by removing cycles) – Any shortest-path in graph G can be no longer than n – 1 edges, where n is the number of vertices

February 8, Relaxation For each vertex in the graph, we maintain d[v], the estimate of the shortest path from s, initialized to at start Relaxing an edge (u,v) means testing whether we can improve the shortest path to v found so far by going through u uv vu 2 2 Relax(u,v) uv vu 2 2 Relax(u,v) Relax (u,v,w) if d[v] > d[u]+w(u,v)then d[v] d[u]+w(u,v) [v] u

February 8, Dijkstra's Algorithm Non-negative edge weights Greedy, similar to Prim's algorithm for MST Like breadth-first search (if all weights = 1, one can simply use BFS) Use Q, priority queue keyed by d[v] (BFS used FIFO queue, here we use a PQ, which is re-organized whenever some d decreases) Basic idea – maintain a set S of solved vertices – at each step select "closest" vertex u, add it to S, and relax all edges from u

February 8, Dijkstras Pseudo Code Graph G, weight function w, root s relaxing edges

February 8, Dijkstras Example s uv yx s uv yx uv s yx s uv yx

February 8, Observe – relaxation step (lines 10-11) – setting d[v] updates Q (needs Decrease-Key) – similar to Prim's MST algorithm Dijkstras Example (2) uv yx uv yx

February 8, Dijkstras Correctness We will prove that whenever u is added to S, d[u] = (s,u), i.e., that d is minimum, and that equality is maintained thereafter Proof – Note that v, d[v] (s,v) – Let u be the first vertex picked such that there is a shorter path than d[u], i.e., that d[u] (s,u) – We will show that this assumption leads to a contradiction

February 8, Dijkstra Correctness (2) Let y be the first vertex V – S on the actual shortest path from s to u, then it must be that d[y] = (s,y) because – d[x] is set correctly for y's predecessor x S on the shortest path (by choice of u as the first vertex for which d is set incorrectly) – when the algorithm inserted x into S, it relaxed the edge (x,y), assigning d[y] the correct value

February 8, But d[u] > d[y] algorithm would have chosen y (from the PQ) to process next, not u Contradiction Thus d[u] = (s,u) at time of insertion of u into S, and Dijkstra's algorithm is correct Dijkstra Correctness (3)

February 8, Dijkstras Running Time Extract-Min executed |V| time Decrease-Key executed |E| time Time = |V| T Extract-Min + |E| T Decrease-Key T depends on different Q implementations QT(Extract -Min) T(Decrease- Key) Total array (V) (1) (V 2 ) binary heap (lg V) (E lg V) Fibonacci heap (lg V) (1) (amort.) (V lgV + E)

February 8, Bellman-Ford Algorithm Dijkstras doesnt work when there are negative edges: – Intuition – we can not be greedy any more on the assumption that the lengths of paths will only increase in the future Bellman-Ford algorithm detects negative cycles (returns false) or returns the shortest path-tree

February 8, Bellman-Ford Algorithm Bellman-Ford(G,w,s) 01 for each v V[G] 02 d[v] 03 d[s] 0 04 [s] NIL 05 for i 1 to |V[G]|-1 do 06 for each edge (u,v) E[G] do 07 Relax (u,v,w) 08 for each edge (u,v) E[G] do 09 if d[v] > d[u] + w(u,v) then return false 10 return true

February 8, Bellman-Ford Example 5 s zy x t -4 s zy x t -4 s zy x t -4 s zy x t

February 8, Bellman-Ford Example s zy x t -4 Bellman-Ford running time: – (|V|-1)|E| + |E| = (VE) 5

February 8, Correctness of Bellman-Ford Let i (s,u) denote the length of path from s to u, that is shortest among all paths, that contain at most i edges Prove by induction that d[u]= i (s,u) after the i-th iteration of Bellman-Ford – Base case (i=0) trivial – Inductive step (say d[u] = i-1 (s,u)): Either i (s,u) = i-1 (s,u) Or i (s,u) = i-1 (s,z) + w(z,u) In an iteration we try to relax each edge ((z,u) also), so we will catch both cases, thus d[u] = i (s,u)

February 8, Correctness of Bellman-Ford After n-1 iterations, d[u] = n-1 (s,u), for each vertex u. If there is still some edge to relax in the graph, then there is a vertex u, such that n (s,u) < n- 1 (s,u). But there are only n vertices in G – we have a cycle, and it must be negative. Otherwise, d[u]= n-1 (s,u) = (s,u), for all u, since any shortest path will have at most n-1 edges

February 8, Shortest-Path in DAGs Finding shortest paths in DAGs is much easier, because it is easy to find an order in which to do relaxations – Topological sorting! DAG-Shortest-Paths(G,w,s) 01 for each v V[G] 02 d[v] 03 d[s] 0 04 topologically sort V[G] 05 for each vertex u, taken in topolog. order do 06 for each vertex v Adj[u] do 07 Relax(u,v,w)